Difference between revisions of "2008 AMC 8 Problems/Problem 3"
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==Solution== | ==Solution== | ||
We can go backwards by days, but we can also backwards by weeks. If we go backwards by weeks, we see that February 6 is a Friday. If we now go backwards by days, February 1 is a <math>\boxed{\textbf{(A)}\ \text{Sunday}}</math> | We can go backwards by days, but we can also backwards by weeks. If we go backwards by weeks, we see that February 6 is a Friday. If we now go backwards by days, February 1 is a <math>\boxed{\textbf{(A)}\ \text{Sunday}}</math> | ||
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| + | ==Solution 2== | ||
| + | Since the days of the week repeat every 7 days and <math>1-13\equiv2\pmod7</math>, the day of the week of the day 2 days after February 13 is the same as the day of the week of February 1. Since we know that February 13 is a Friday, the day of the week of the day 2 days after February 13 is a Sunday so we get that February 1 is a <math>\boxed{\textbf{(A)}\ \text{Sunday}}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2008|num-b=2|num-a=4}} | {{AMC8 box|year=2008|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 16:44, 22 October 2020
Contents
Problem
If February is a month that contains Friday the 
, what day of the week is February 1?
Solution
We can go backwards by days, but we can also backwards by weeks. If we go backwards by weeks, we see that February 6 is a Friday. If we now go backwards by days, February 1 is a 
Solution 2
Since the days of the week repeat every 7 days and 
, the day of the week of the day 2 days after February 13 is the same as the day of the week of February 1. Since we know that February 13 is a Friday, the day of the week of the day 2 days after February 13 is a Sunday so we get that February 1 is a .
See Also
| 2008 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
